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Math Help - Absolute Convergence of Integral

  1. #1
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    Absolute Convergence of Integral

    I'm having a hard time showing that \lim_{c \to \0} \int_{c}^{1}|\frac{sin(\frac{1}{x})}{x}|dx fails to converge absolutely, any insight would be greatly appreciated.
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  2. #2
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    Try writing \displaystyle \sin{\left(\frac{1}{x}\right)} as a Taylor Series, then simplifying \displaystyle \left|\frac{\sin{\left(\frac{1}{x}\right)}}{x} \right|.
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  3. #3
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    Another way: If you know that  \int_1^{\infty} \left| \frac{\sin (z)}{z} \right|dz is infinite, just apply a change of variables.
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    Thanks

    Thanks to both of you for your help. I did try one way before I read this, maybe you can tell me this works...

    \lim_{c\to\0}\int_{c}^{1}|\frac{sin(\frac{1}{x})}{  x}|dx\geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }|\frac{sin(\frac{1}{x})}{x}|dx \geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }\frac{dx}{x}

    ...and then just evaluate, I could be way off.
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  5. #5
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    Quote Originally Posted by eastman View Post
    Thanks to both of you for your help. I did try one way before I read this, maybe you can tell me this works...

    \lim_{c\to\0}\int_{c}^{1}|\frac{sin(\frac{1}{x})}{  x}|dx\geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }|\frac{sin(\frac{1}{x})}{x}|dx \geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }\frac{dx}{x}

    ...and then just evaluate, I could be way off.
    How did you obtain the last inequality? There is alway a zero in those intervals so the elementary minimum bound doesn't work here.
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    Sorry, I'm slightly confused, could you elaborate?
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  7. #7
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    Disregard the second part of my last message, could you explain how you got the last inequality?
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  8. #8
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    You're right the last inequality does not hold, sin vanishes over those intervals.. I ended up doing it with the change of variables like you orginally said. Thanks for your help.
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